[simpits-tech] electrical ??
Heidner, Troy W
simpits-tech@simpits.org
Mon, 16 Sep 2002 12:00:37 -0500
Dave,
I just finished reading all the posts in this thread. I decided that I have
a couple of comments that might be helpful and haven't been specifically
addressed yet. I'm an electrical engineer. I thought I might offer a
couple of insights.
The current rating of the individual bulb gives you most of the information
that you need, but you also need to know if the lighting circuit that the
bulbs are used in is wired in series, parallel or some combination thereof.
Most likely it is wired entirely in parallel. You can pretty much determine
that from the facts. You said the supply voltage was 5V and the bulbs are
rated for 5V. The only way to get 5V to each bulb with a 5V supply is to
wire them in parallel. That being the case, then your calculation (0.115A x
10 bulbs = 1.15A) is correct.
As a side note, in case you're interested or encounter different wiring,
here are two other possibilities. If they were wired in series then the
total current draw would be equal to the current draw of a single bulb
(0.115A). Sometimes lighting circuits are wired in a combination of series
and parallel. It's done sometimes to simplify wiring and sometimes to meet
certain voltage requirements. (I used to be an avionics tech in the Air
Force and I remember some panel lighting like this. Mostly in older
aircraft.) You can tell if there is any series wired lighting by removing
single bulbs. If any other bulbs go out when you pull any single bulb, all
of those bulbs are on the same series circuit. To find the total current
draw in a situation like this you need to find the total number of series
bulb groups and then multiply that number by the current draw of a single
bulb. Each series group draws the same current as a single bulb. If that
doesn't make sense I wouldn't worry about it, it sounds like you are dealing
with a purely parallel circuit anyway.
I have two other comments that might be helpful. One has to do with making
current measurements with a multimeter. I taught electronics for about 5
years at a community college. I saw more multimeters with blown fuses from
incorrectly made current measurements that you would believe. The good news
is that all multimeters have fuses to protect them from too much current in
a current measurement so you shouldn't be able to hurt it even if you do it
wrong (but if after making a measurement you find that you're reading 0 Amps
all the time you might check the fuse). The current measurement in a
multimeter is actually a separate circuit than the one for voltage and
resistance measurements. You can blow the current measurement fuse and
still make voltage and resistance measurements with your meter. Anyway, the
natural tendency when using a meter is to first select the meter function
and scale that you want and then just put your test probes ACROSS the
terminals or device that you want to measure the current through. That is
***NOT*** the correct way to do it. That IS the correct way to blow a fuse
in your meter though, if such is your objective. :-) As Roy stated in a
reply, you must make current measurements in series with the object you want
to measure the current flow in. That means that you actually have break the
circuit somewhere and insert the meter IN LINE with the device to be
measured. Not ACROSS. Usually the easiest way to do that would be at the
power supply. Just hook one side (it doesn't matter which, let's make it
the negative side for reference purposes) of the power source up to your
light circuit as normal. Then hook the red lead of your meter to the
positive side of the power source and hook your black test lead to the other
contact point of your lighting circuit. That should measure the total
current draw. One last thing to be aware of with current measurements.
Some meters have very small current measurement capabilities. 200mA is
common. Your circuit probably draws in the neighborhood of 1.15A or 1150mA.
Obviously that won't work. Also, most (but not all) multimeters with
current measurement capabilities over 200mA have a separate test lead
connection point for higher current measurements. Just make sure you have
your test lead plugged into the right hole!
Okay last point. (Sorry if this is dragging on.) This has to do with
resistance measurements and the application of Ohm's Law. First of all,
bravo to you, and all the other message repliers, everyone has made correct
theoretical applications of Ohm's law in all it's forms! But... There is a
But... There is an application here that is still incorrect in practice.
You stated that you measured the resistance of the circuit and it came out
to be around .5 Ohms. That was probably a correct measurement (or close
anyway, many multimeters aren't tremendously accurate at resistance ranges
that small). With that correct measurement you also made a correct
calculation of 10 amps of current draw. You said that sounded high. Well
it should sound high. We have already established via specifications and
circuit deductions that the current draw is more like 1.15 amps. Where is
the problem here? I'll tell you. The filament resistance of an
incandescent light bulb goes way up when the light bulb is on because the
heat of the filament is so much greater. The resistance of a typical low
voltage incandescent light bulb is around 50 ohms when it's lit. In fact
from the specifications of this bulb, you can calculate the approximate
resistance that they should have when lit. According to Ohm's law.
Resistance = Voltage / Current. From the specs. this works out to be
5V/0.115A = 43.5 Ohms. Then when you put ten 43.5 Ohm bulbs in parallel the
total circuit resistance becomes 4.35 Ohms. Then using Ohm's law again
(just to validate) Current = Voltage / Resistance 5V / 4.35 Ohms = 1.15
Amps. Total current flow. So, that's where the trouble is. Your
resistance measurement is made with the bulbs off and the resistance is very
different with the bulbs on. But don't try to measure the resistance with
your meter with power applied to the circuit! Multimeters are not designed
to be able to make resistance measurements that way.
Well, that's all I have to add. I hope it was helpful. I also hope it
wasn't too long-winded. I've been accused of that before! ;-) My wife
says it's the teacher in me. ;-) If you have any other questions that I
could help with, I'd be happy to answer.
Troy
> -----Original Message-----
> From: Dave Hensley [mailto:dphens@nc.rr.com]
> Sent: Saturday, September 14, 2002 1:57 PM
> To: simpits-tech@simpits.org
> Subject: Re: [simpits-tech] electrical ??
>
>
> Thanks! I did find this, which may help. On the back of the
> panel it has
> "10#7152 LAMPS". I'm assuming this means there are 10 lamps
> in the panel,
> part number 7152. I searched on this and found a PDF doc with
> lamp part
> numbers.
>
> Line Part
> Filament
> Life
>
> No. No. Volts Amps M.S.C.P. Type
> Hours
>
> 7 7152 5.0 0.115 0.15 C-2R 40,000
>
>
> So, IF that is the lamp being used then could I figure the amps by
> 0.115*10(number of lamps)=1.15. I'm starting to feel in over my head a
> little, so I may do some studying and come back to this later. I still
> appreciate any help you guys can give.
>
> Dave
>
> ----- Original Message -----
> From: "Andreas Fransson" <andreas.fransson@post.utfors.se>
> To: <simpits-tech@simpits.org>
> Sent: Saturday, September 14, 2002 2:05 PM
> Subject: Re: [simpits-tech] electrical ??
>
>
> > (disclamer: my high school electrical skills may be a bit rusty...)
> >
> > IIRC...
> >
> > P=UI (watts = volts x amps)
> > I=U/R (amps = volts / ohms)
> > =>
> > P=U^2/R (watts = volts(square) / ohms)
> >
> > You have already concluded that the panel has a 0.5 ohm
> resistance, and is
> > rated at 5 volts. This means that it draws 5^2/0.5=50 watts at that
> voltage.
> > That seems to me to be a whole lot...
> >
> > At 5 volts, the current should be 5/0.5=10 amps...
> >
> > Someone please correct me if I have gotten all of this backwards. ;)
> >
> > Andreas
> > http://valhallainc.d2g.com
> >
> >
> > From: "Dave Hensley" <dphens@nc.rr.com>
> > > I need a little assistance with a power problem. I am trying to
> determine
> > > the amount of amps I need going through the panels I have
> to light them
> > up.
> > > The panels are rated for 5vdc. I plugged in a 6vdc
> adapter which is
> > putting
> > > out 7.70v before anything is connected, measured with a
> mutlimeter. Then
> I
> > > connected leads to the adapter and the panel, while
> leaving the leads
> for
> > > the multimeter connected also. The volts in the
> multimeter dropped to
> > 4.41v.
> > > Next I connected only the meter to the panel set to
> measure resistance
> and
> > > had a reading that fluctuated between .4 to .6, so I split the
> difference
> > at
> > > .5. So, using V=I*R formula, 7.70/.5=3.85. So, the lights
> are pulling
> > almost
> > > 4 amps? Am I even close to having that right?
> > >
> > > Obviously you guys are seeing that I know very little
> about this and am
> in
> > > the process of learning. I am playing with the panel in
> an attempt to
> > > understand the relationship of volts, current and
> resistance. If am I
> way
> > > off at this point, I'll stop and do some more reading. I
> am just hoping
> I
> > am
> > > on the right track. :-)
> > >
> > > Thanks!
> > >
> > > Dave
> >
> >
> >
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>
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